{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# hw8 - Killough Ian Homework 8 Due 4:00 pm Inst Donna C Lyon...

This preview shows pages 1–3. Sign up to view the full content.

Killough, Ian – Homework 8 – Due: Nov 14 2006, 4:00 pm – Inst: Donna C Lyon 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. This HW assignment is due Tuesday, November 14, by 4:00PM. 001 (part 1 of 3) 10 points Make the following assumptions: 1) The density of liquid water at 100 C is 1.00 g/mL; 2) the vapor is adequately described by the ideal gas equation; 3) the external pressure is constant at 1.00 atm; and 4) the heat of vaporization of water is 2.26 kJ/g. What is q for the vaporization of 1.00 g of liquid water at 1.00 atm and 100 C, to the formation of 1.00 g of steam at 1.00 atm and 100 C? 1. q = 3 . 24 kJ 2. q = 0 . 126 kJ 3. q = 0 kJ 4. q = 1 . 67 kJ 5. q = 2 . 26 kJ correct Explanation: m = 1.00 g Δ H vap . = 2.26 kJ/g q = 2 . 26 kJ g (1 g) = 2 . 26 kJ 002 (part 2 of 3) 10 points What is w for this process? 1. w = - 1 . 70 J 2. w = - 1 . 70 L · atm correct 3. w = +1 . 70 L · atm 4. w = +1 . 70 J Explanation: P = 1 atm T = 100 C + 273 = 373 K density = 1 g/mL W = - P Δ V V i = (1 g) 1 mL 1 g 1 L 1000 mL · = 0 . 001 L V f = n R T P = 0 . 082058 L · atm 1 mol · K 373 K atm × (1 g) 1 mol 18 g = 1 . 701 L w = - (1 atm)(1 . 701 - 0 . 001) L = - 1 . 70 L · atm 003 (part 3 of 3) 10 points What is Δ E for this process? 1. Δ E = 2 . 43 kJ 2. Δ E = 2 . 09 kJ correct 3. Δ E = 0 . 56 kJ 4. Δ E = 1 . 29 kJ Explanation: Δ E = q + w = 2 . 26 kJ + - 1 . 70 L · atm · × 101 . 33 J L · atm ·‡ 1 kJ 1000 J · = 2 . 09 kJ 004 (part 1 of 1) 10 points Calculate the quantity of energy required to change 3.00 mol of liquid water at 100 C to steam at 100 C. The molar heat of vaporiza- tion of water is 40.6 kJ/mol. 1. 13.5 kJ 2. 40.6 kJ 3. 122 kJ correct

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Killough, Ian – Homework 8 – Due: Nov 14 2006, 4:00 pm – Inst: Donna C Lyon 2 4. None of these 5. 300 kJ Explanation: n = 3.0 mol 40 . 6 kJ mol · 3 mol = 121 . 8 kJ 005 (part 1 of 1) 10 points How much heat is required to vaporize 50.0 g of water if the initial temperature of the water is 25.0 C and the water is heated to its boiling point where it is converted to steam? The specific heat capacity of water is 4 . 18 J · ( C) - 1 · g - 1 and the standard en- thalpy of vaporization of water at its boiling point is 40 . 7 kJ · mol - 1 . 1. 23.5 kJ 2. 64.2 kJ 3. 40.7 kJ 4. 129 kJ correct 5. 169 kJ Explanation: m = 50 g T i = 25 C T f = 100 C C = 4 . 18 J / g / C Δ H vap = 40 . 7 kJ/mol H = m C Δ T + m MM Δ H vap = (50 g) (4 . 18 J / g / C) × (100 - 25) C 1 kJ 1000 J + 50 g 18 g / mol (40 . 7 kJ / mol) = 128 . 731 kJ 006 (part 1 of 1) 10 points Calculate the standard reaction enthalpy for the reaction.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

hw8 - Killough Ian Homework 8 Due 4:00 pm Inst Donna C Lyon...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online